Problem 21 - Applied Calculus

Question

Estimate the instantaneous rate of change of the function \(f(x)=x \ln x\) at \(x=1\) and at \(x=2 .\) What do these values suggest about the concavity of the graph between 1 and \(2 ?\)

Step-by-Step Solution

Answer

At \(x=1\), the rate of change is 1, and at \(x=2\), it is approximately 1.693. The graph is concave up between 1 and 2.

Step 1: Find the Derivative

The first step in estimating the instantaneous rate of change of the function is to find its derivative. The derivative of a function gives the rate at which the function is changing at any point on the graph.For the function \(f(x) = x \ln x\), we use the product rule, which states that the derivative of a product \(uv\) is \(u'v + uv'\). Here, \(u = x\) and \(v = \ln x\). Thus,\[ \frac{d}{dx}[x \ln x] = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1. \]

Step 2: Calculate the Derivative at x=1

Now we find the derivative of the function at \(x=1\) using the expression derived in Step 1.\[ f'(1) = \ln(1) + 1 = 0 + 1 = 1. \]

Step 3: Calculate the Derivative at x=2

Substitute \(x=2\) into the derivative to find the instantaneous rate of change at this point.\[ f'(2) = \ln(2) + 1. \]Using the known value \(\ln(2) \approx 0.693\), we get\[ f'(2) \approx 0.693 + 1 = 1.693. \]

Step 4: Analyze Concavity

To determine the concavity between \(x = 1\) and \(x = 2\), we look at the sign of the second derivative in that interval. Calculate the second derivative by differentiating the first derivative, \(f'(x) = \ln x + 1\).\[ f''(x) = \frac{d}{dx}(\ln x + 1) = \frac{1}{x}. \]Since \(f''(x) = \frac{1}{x}\) is positive for \(x > 0\), the function \(f(x) = x \ln x\) is concave up on the interval from \(1\) to \(2\).

Key Concepts

Derivative

When we talk about the instantaneous rate of change of a function, what we're referring to is the derivative. The derivative of a function at a particular point tells us how quickly the function's value is changing at that exact point.
For the function given in the exercise, \( f(x) = x \ln x \), we are interested in finding this rate at certain points, specifically at \( x = 1 \) and \( x = 2 \).
To find the derivative, we employ a fundamental tool in calculus called the product rule.

• The product rule is used when you need to differentiate a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), and their product is \( uv \), then the derivative is given by \( u'v + uv' \).
• In our case, \( u(x) = x \) and \( v(x) = \ln x \), so applying the product rule results in \( f'(x) = \ln x + 1 \).

Once we have our derivative, calculating it at particular points (like \( x = 1 \) or \( x = 2 \)) gives us the instantaneous rates of change at those points.

Concavity

Concavity helps us understand the direction in which a curve is bending. Whether a function is bending upwards or downwards is determined by studying its second derivative. In these terms:

• If the second derivative, \( f''(x) \), is positive over an interval, the function is said to be concave up, resembling a U-shape.
• If \( f''(x) \) is negative, the function is concave down, resembling an upside-down U.

For our function, the second derivative is \( f''(x) = \frac{1}{x} \), which is always positive for \( x > 0 \).
Therefore, between the points \( x = 1 \) and \( x = 2 \), the function \( f(x) = x \ln x \) is concave up. This means the graph is bending upwards along this interval.

Product Rule

The product rule is essential when differentiating products of two functions. Here's a breakdown of how and when to use it:

• It's crucial whenever you're faced with a differentiation task involving two functions multiplied together. Simply doing the derivative of each part separately won't work.
• Think of it like this: if you have a product \( uv \), the derivative becomes \( u'v + uv' \). This means you differentiate \( u \), keep \( v \) as it is, and vice versa, then sum the two results.

In the context of the example \( f(x) = x \ln x \), the product rule was used to effectively split the work between the two parts, \( x \) and \( \ln x \).
Applying the rule ensured that we correctly derived the expression \( \ln x + 1 \) for the first derivative, capturing how both the \( x \) and the \( \ln x \) components contribute to the change at any given point.